Show the torus of revolution has no umbilic points

Question

Let $\mathbb{T}$ denote the torus of revolution with the usual parametrization: $x(u,v) = ( (R + r\cos(u))\cos(v), (R + r\cos(u))\sin(v), r\sin(u) )$

Show that $\mathbb{T}$ has no umbilic points.

Answer 1

Simple calculation shows that $$x_u = ( -r\sin(u)\cos(v), -r\sin(u)\sin(v), r\cos(u)),$$ $$x_v = ( -(R + r\cos(u))\sin(v), (R + r\cos(u))\cos(v), 0),$$ $$x_{uu} = ( -r\cos(u)\cos(v), -r\cos(u)\sin(v), -r\sin(u)),$$ $$x_{uv} = ( r\sin(u))\sin(v), -r\sin(u))\cos(v), 0),$$ $$x_{vv} = ( -(R + r\cos(u))\cos(v), -(R + r\cos(u))\sin(v), 0).$$ These imply that the coefficients of the first fundamental form are $$E=x_u\cdot x_u=r^2, F=x_u\cdot x_v=0, G=x_v\cdot x_v=(R + r\cos(u))^2,$$ and the unit normal is given by $$n=\frac{u\times v}{\|u\times v\|}=( -\cos(u)\cos(v), -\cos(u)\sin(v), -\sin(u)).$$ Hence, the coefficients of the second fundamental form are $$e=x_{uu}\cdot n=r, f=x_{uv}\cdot n=0, g=x_{vv}\cdot n=(R + r\cos(u))\cos(u).$$ Therefore, the Gauss curvature $K$ and mean curvature $H$ are given by $$K=\frac{eg-f^2}{EG-F^2}=\frac{\cos(u)}{r(R + r\cos(u))}, H=\frac{eG+Eg-2fF}{EG-F^2}=\frac{R + 2r\cos(u)}{r(R + r\cos(u))}.$$

Now note that the principal curvatures $\kappa_1, \kappa_2$ are the roots of the quadratic equation $$\kappa^2-2H\kappa+K=0,$$ and by definition a point $x$ is umbilic if $\kappa_1=\kappa_2$. Therefore, $x$ is umbilic if and only if the above quadratic equation has double roots, which is equivalent to the discriminant $(-2H)^2-4K=0$, i.e. $H^2=K$. However, by using the above expressions for $K$ and $H$, we see that $$H^2-K=\frac{(R + 2r\cos(u))^2}{r^2(R + r\cos(u))^2}-\frac{\cos(u)}{r(R + r\cos(u))}=\frac{\frac{R^2}{4}+3(\frac{R}{2}+r\cos(u))^2}{r^2(R + r\cos(u))^2}>0.$$ This proves there does not exist $x$ such that $\kappa_1=\kappa_2$, i.e. $\mathbb{T}$ has no umbilic points.